∫ e^2x ≤ft(sin x +1/2 cos x) dx is equal to

The correct answer is Option (2) → \(\displaystyle\frac{1}{2}e^{2x}\sin x+C\), C is an arbitrary constant

Taking \(\frac{1}{2}\) out of the bracket:

    \( \displaystyle e^{2x}\left(\sin x + \frac{1}{2}\cos x\right) = \frac{1}{2}\,e^{2x}(2\sin x + \cos x) \)

By the product rule,

    \( \displaystyle \frac{d}{dx}\left(e^{2x}\sin x\right) = 2e^{2x}\sin x + e^{2x}\cos x \)

    \(= e^{2x}(2\sin x + \cos x)\)

So the bracket is exactly the derivative of \(e^{2x}\sin x\). This is the standard form \(\int e^{ax}\left[a f(x) + f'(x)\right]dx = e^{ax}f(x) + C\), with \(a = 2\) and \(f(x) = \sin x\):

    \( \displaystyle \int e^{2x}\left(\sin x + \frac{1}{2}\cos x\right)dx = \frac{1}{2}\int e^{2x}(2\sin x + \cos x)\,dx \)

    \( \displaystyle = \frac{1}{2}\,e^{2x}\sin x + C \)